'''
根据 逆波兰表示法，求表达式的值。

有效的算符包括 +、-、*、/ 。每个运算对象可以是整数，也可以是另一个逆波兰表达式。

注意 两个整数之间的除法只保留整数部分。

可以保证给定的逆波兰表达式总是有效的。换句话说，表达式总会得出有效数值且不存在除数为 0 的情况。

 

示例 1：

输入：tokens = ["2","1","+","3","*"]
输出：9
解释：该算式转化为常见的中缀算术表达式为：((2 + 1) * 3) = 9
示例 2：

输入：tokens = ["4","13","5","/","+"]
输出：6
解释：该算式转化为常见的中缀算术表达式为：(4 + (13 / 5)) = 6
示例 3：

输入：tokens = ["10","6","9","3","+","-11","*","/","*","17","+","5","+"]
输出：22
解释：该算式转化为常见的中缀算术表达式为：
  ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/evaluate-reverse-polish-notation
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
'''
class Solution(object):
    def evalRPN(self, tokens):
        """
        :type tokens: List[str]
        :rtype: int
        """
        stack = []
        for token in tokens:
            try:
                stack.append(int(token))
            except:
                num2 = stack.pop()
                num1 = stack.pop()
                stack.append(self.evaluate(num1, num2, token))
        return stack[0]

    def evaluate(self, num1, num2, op):
        if op == "+":
            return num1 + num2
        elif op == "-":
            return num1 - num2
        elif op == "*":
            return num1 * num2
        elif op == "/":
            return int(num1 / float(num2))

tokens = ["10","6","9","3","+","-11","*","/","*","17","+","5","+"]
Solution().evalRPN(tokens)